Can someone give me the answers

Which of the following correctly describe electric field lines?

Xelga [282]

-- Electric field lines DO never cross. <em>(A) </em>

-- Electric field lines that are close together DO indicate a stronger electric field. <em>(B) </em>

-- Electric field lines DO not affect the charge that created them. <em>(C)</em>

-- Electric field lines DON'T begin on north poles and end on south poles. North and South "poles" are the way we talk about magnets, not electric charges.

6 0

3 years ago

Read 2 more answers

When you hear an ambulance siren, it alternates between high and low tones, depending on the frequency of sound waves. this is c

dangina [55]

<span>Frequency of a sound wave is called the pitch. Higher frequencies have a higher pitch and lower frequencies have the opposite. When an ambulance travels by a listener, the frequencies are oscillating rapidly and causing the shrill, loud sounds that emanate from the sirens.</span>

6 0

3 years ago

A parallel plate air capacitor has a capacitance of 10 to the power -9. What potential difference is required for a charge of 5×

Maurinko [17]

The potential difference across the capacitor is 5 × 10∧4 volts and the energy stored in it is 1. 25 Joules

<h3>What is the energy in a capacitor?</h3>

The energy stored in a capacitor is an electrostatic potential energy.

It is related to the charge(Q) and voltage (V) between the capacitor plates.

It is represented as 'U'.

<h3>How to determine the potential difference</h3>

Formula:

Potential difference, V is the ratio of the charge to the capacitance of a capacitor.

It is calculated using:

V = Q ÷ C

Where Q = charge 5 × 10∧-5C and C = capacitance 10∧-9

Substitute the values into the equation

Potential difference, V = 5 × 10∧-5 ÷ 10∧-9 = 5 × 10∧4 volts

<h3>How to determine the energy stored</h3>

Formula:

Energy, U = 1 ÷ 2 (QV)

Where Q= charge and V = potential difference across the capacitor

Energy, U = 1 ÷ 2 ( 5 × 10∧-5 × 5 × 10∧4)

= 0.5 × 25 × 10∧-1

= 0.5 × 2.5

= 1. 25 Joules

Therefore, the potential difference across the capacitor is 5 × 10∧4 volts and the energy stored in it is 1. 25 Joules

Learn more about capacitance here:

brainly.com/question/14883923

#SPJ1

6 0

1 year ago

Coherent light of wavelength 525 nm passes through two thin slits that are 4.15Ã—10^(âˆ’2) mm apart and then falls on a screen 7

IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

$sin \theta=\frac{n \lambda}{d}$

where

n is the order of the maximum

$\lambda$ is the wavelength

$a$ is the distance between the slits

In this problem,

n = 5

$\lambda=525 nm =5.25\cdot 10^{-7} m$

$a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m$

So we find

$\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}$

And given the distance of the screen from the slits,

$D=75.0 cm = 0.75 m$

The distance of the 5th bright fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm$

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

$sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}$

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

$\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}$

And the distance of the 8th dark fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm$

5 0

3 years ago

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

<em>c) </em> $\frac{4492.18}{v_{car} ^{2} }$

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = <em>6738.27 J</em>

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s